zoj 3814 Sawtooth Puzzle(隐式图搜索)

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题目链接：zoj 3814 Sawtooth Puzzle

题目大意：给定一个9宫拼图，每次可以挑选一个位置顺时针旋转，和普通拼图不一样的是每块拼图周围可能有齿转动一个可能导致全部拼图转变。

解题思路：隐式图搜索，9块拼图最多49个状态，对于每个状态枚举转动的位置，考虑转动的状态。一开始转移是用bfs写的，但是由于频繁申请队列，然后时间爆了

#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#include <algorithm>

using namespace std;
typedef long long ll;
typedef char Mat[10][10];

const int maxn = 9;
const int maxm = 4;
const int maxs = 1<<18;
const int dir[4][2] = { {0, -1}, {-1, 0}, {0, 1}, {1, 0} };

Mat sb[maxn], se[maxn];
int mesh[maxn][maxm];

inline int hash_state(int* c) {
    int ret = 0;
    for (int i = 0; i < maxn; i++)
        ret = ret * maxm + c[i];
    return ret;
}

inline void reback(int p, int* c) {
    for (int i = maxn-1; i >= 0; i--) {
        c[i] = p % maxm;
        p /= maxm;
    }
}

void rotate (Mat u);

int vis[maxs+5], c[maxn+5];
set<int> stop;

void dfs (int d, int s) {
    if (d >= maxn) {
        stop.insert(s);
        return;
    }

    for (int i = 0; i < 4; i++) {
        if (memcmp(sb[d], se[d], sizeof(se[d])) == 0)
            dfs(d+1, s * maxm + i);
        rotate(sb[d]);
    }
}

void init () {

    stop.clear();

    for (int i = 0; i < 3; i++) {
        for (int k = 0; k < 8; k++) {
            for (int j = 0; j < 3; j++)
                scanf("%s", sb[i*3+j][k]);
        }
    }

    for (int i = 0; i < 3; i++) {
        for (int k = 0; k < 8; k++) {
            for (int j = 0; j < 3; j++)
                scanf("%s", se[i*3+j][k]);
        }
    }

    dfs(0, 0);

    for (int i = 0; i < maxn; i++) {
        for (int j = 0; j < maxm; j++)
            scanf("%d", &mesh[i][j]);
    }
}

void get(int u, int* t) {

    for (int i = 0; i < 4; i++) {
        int p = u / 3 + dir[i][0];
        int q = u % 3 + dir[i][1];
        int v = p * 3 + q;
        if (p < 0 || p >= 3 || q < 0 || q >= 3 || c[v])
            continue;

       if (mesh[u][(t[u] + i) % maxm] && mesh[v][(t[v] + i + 2) % maxm]) {
           c[v] = -c[u];
           get(v, t);
       }
    }
}

int solve (int* t, int x) {

    memset(c, 0, sizeof(c));
    c[x] = -1;
    get(x, t);

    int ret = 0;
    for (int i = 0; i < maxn; i++)
        ret = ret * 4 + (t[i] + c[i] + maxm) % maxm;
    return ret;
}

int bfs() {

    if (stop.size() == 0)
        return -1;

    queue<int> que;
    memset(vis, -1, sizeof(vis));

    que.push(0);
    vis[0] = 0;

    if (stop.find(0) != stop.end())
        return 0;

    int t[maxn+5];

    while (!que.empty()) {
        int u = que.front();
        que.pop();

        reback(u, t);

        for (int i = 0; i < maxn; i++) {
            int v = solve(t, i);

            if (vis[v] >= 0)
                continue;
            vis[v] = vis[u] + 1;
            que.push(v);

            if (stop.find(v) != stop.end())
                return vis[v];
        }
    }
    return -1;
}

int main () {
    int cas;
    scanf("%d", &cas);

    while (cas--) {
        init();
        printf("%d\n", bfs());
    }
    return 0;
}

void rotate (Mat u) {
    Mat v;
    memset(v, 0, sizeof(v));
    for (int i = 0; i < 8; i++) {
        for (int j = 0; j < 8; j++)
            v[7-j][i] = u[i][j];
    }
    memcpy(u, v, sizeof(v));
}

zoj 3814 Sawtooth Puzzle(隐式图搜索)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/39179099